Write the valence-shell electron configurations and bond orders of (a) \(\mathrm{B}_{2} ;\) (b) \(\mathrm{Be}_{2} ;\) (c) \(\mathrm{F}_{2}\).

Short Answer

Expert verified
The valence-shell electron configurations and bond orders are: (a) B2: Bond order 1, (b) Be2: Bond order 0 (nonexistent molecule), (c) F2: Bond order 1.

Step by step solution

01

Valence-shell electron configuration for B2

To find the valence electron configuration for a B2 molecule, start by writing the electron configuration for a single Boron (B) atom, which is 1s2 2s2 2p1. Since there are two Boron atoms in B2, combine their valence electrons in molecular orbitals. B2 has a total of 6 valence electrons to place in the molecular orbitals. The valence orbitals fill from lowest to highest energy level: sigma2s, sigma*2s, pi2p (two orbitals degenerate), and sigma2p. For B2, the filling order is sigma2s(2), sigma*2s(2), pi2p(2), giving a bond order of 1.
02

Valence-shell electron configuration for Be2

A single Beryllium (Be) atom has an electron configuration of 1s2 2s2. Be2 therefore has a total of 4 valence electrons. These electrons will fill the molecular orbitals as: sigma2s(2), sigma*2s(2). As the sigma and sigma* molecular orbitals are fully filled, they cancel each other out, leading to no net bonding interactions. Hence, the bond order for Be2 is 0, which indicates that a Be2 molecule does not exist under normal conditions.
03

Valence-shell electron configuration for F2

For a Fluorine (F) atom, the electron configuration ends with 2s2 2p5. In F2, there are a total of 14 valence electrons. For the molecular orbitals, the electrons will fill as: sigma2s(2), sigma*2s(2), pi2p(4), sigma2p(2), pi*2p(4). Only 2 electrons are in the bonding orbitals not cancelled out by electrons in antibonding orbitals, resulting in a bond order of 1.
04

Calculate bond orders

The bond order is half the difference between the number of bonding and antibonding electrons. For B2, bond order = (4 - 2) / 2 = 1. For Be2, bond order = (2 - 2) / 2 = 0. For F2, bond order = (8 - 6) / 2 = 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Orbital Theory
Understanding molecular orbital theory is crucial when delving into the intricacies of chemical bonding. It is a fundamental concept used to explain the bonding between atoms within a molecule.

At its core, molecular orbital theory describes how atomic orbitals combine to form molecular orbitals when two atoms bond. These molecular orbitals can be bonding or antibonding, with bonding orbitals stabilizing the molecule and antibonding orbitals destabilizing it. A bonding molecular orbital is lower in energy than the atomic orbitals from which it's formed, while an antibonding orbital is higher in energy.

When considering the valence-shell electron configurations of molecules, it is the distribution of these valence electrons across the available molecular orbitals that determines the strength and type of bonds formed. For instance, in the case of a B2 molecule, electrons are shared between B atoms and occupy the bonding and antibonding orbitals. The order in which these molecular orbitals are filled is based on their energy levels, with the lowest energy orbitals filling up first.

The concept of molecular orbitals provides a detailed explanation beyond the traditional Lewis structures, allowing chemists to predict and explain molecular properties with greater accuracy.
Bond Order Calculation
Bond order calculation is a numerical way to express the stability of a molecule compared to its constituent atoms. It is interpreted as the number of chemical bonds between a pair of atoms and offers insight into the bond length and bond energy.

The bond order can be determined by subtracting the number of electrons in antibonding orbitals from the number of electrons in bonding orbitals, then dividing the result by two. For example, the bond order of a B2 molecule can be calculated using the distribution of electrons across its molecular orbitals. We can see that there are four bonding electrons and two antibonding electrons, resulting in a bond order of 1. The same calculation for F2 also results in a bond order of 1, whereas Be2 has a bond order of 0, indicating no stable bond.

Importantly, a non-zero bond order suggests a stable molecule, with higher order values corresponding to stronger, shorter bonds. Conversely, a bond order of zero implies that the molecule is not stable and does not exist under normal circumstances—as in the case of Be2.
Electron Configuration
Electron configuration plays a pivotal role in understanding the behavior of atoms during chemical bonding. It is the arrangement of electrons in the orbitals of an atom, and the valence electrons (those in the outermost shell) are the ones involved in forming bonds.

The electron configuration for an individual atom is determined by principles such as the Aufbau principle, Hund's rule, and the Pauli exclusion principle. These guide the placement of electrons into the orbitals, filling from the lowest to the highest energies. In molecules, the electron configuration extends to describe how electrons occupy both the bonding and antibonding molecular orbitals.

For example, the electron configuration of boron, 1s2 2s2 2p1, leads to the formation of three molecular orbitals when two boron atoms bond to form B2. By understanding these configurations and how electrons are shared in molecules like F2, chemists can predict magnetic properties, reactivity, and other chemical characteristics that arise from the unique electronic arrangements.

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Most popular questions from this chapter

Consider the bonding in \(\mathrm{CH}_{2}=\mathrm{CHCHO}\). (a) Draw the most important Lewis structure. Include all nonzero formal charges. (b) Identify the composition of the bonds and the hybridization of each lone pair-for example, by writing \(\sigma\left(H 1 s, C 2 \mathrm{sp}^{2}\right)\).

Write its Lewis structure and predict whether each of the following molecules is polar or nonpolar: (a) \(\mathrm{H}_{2} \mathrm{Se}\); (b) \(\mathrm{AsF}_{5}\); (c) \(\mathrm{SiO}_{2} ;\) (d) \(\mathrm{NF}_{3}\).

Predict whether each of the following molecules is likely to be polar or nonpolar: (a) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (pyridine, a molecule like benzene except that one \(-\mathrm{CH}-\) group is replaced by a nitrogen atom); (b) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (ethane); (c) \(\mathrm{CHCl}_{3}\) (trichloromethane, also known as chloroform, a common organic solvent and once used as an anesthetic).

Consider the molecules \(\mathrm{H}_{2} \mathrm{CCH}_{2}, \mathrm{H}_{2} \mathrm{CCCH}_{2}\), and \(\mathrm{H}_{2} \mathrm{CCCCH}_{2}\). (a) Draw Lewis structures for these molecules. (b) What is the hybridization at each \(\mathrm{C}\) atom? (c) What type of bond connects the carbon atoms (single, double, etc.)? (d) What are the \(\mathrm{HCH}, \mathrm{CCH}\), and \(\mathrm{CCC}\) angles in these molecules? (e) Do all the hydrogen atoms lie in the same plane? (f) A generalized formula for molecules of this type is \(\mathrm{H}_{2} \mathrm{C}(\mathrm{C})_{x} \mathrm{CH}_{2}\), where \(x\) is \(0,1,2\), etc. What can be said, if anything, about the relative orientation of the \(\mathrm{H}\) atoms at the ends of the chain as a function of \(x\) ?

Identify the hybrid orbitals used by the phosphorus atom in each of the following species: (a) \(\mathrm{PCl}_{4}^{+}\); (b) \(\mathrm{PCl}_{6}{ }^{-}\); (c) \(\mathrm{PCl}_{5}\); (d) \(\mathrm{PCl}_{3}\).

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